Solution

$[4;\overline{1,3,1,8}]$ is the square root of what value?


Suppose $y= [0;\overline{1,3,1,8}]$. Then,

$$\begin{array}{rcccl} y &=& \frac{1}{1+\frac{1}{3+\frac{1}{1 + \frac{1}{8+y}}}} &=& \frac{1}{1+ \frac{1}{3 + \frac{1}{\frac{9+y}{8+y}}}}\\\\ &=& \frac{1}{1+ \frac{1}{3 + \frac{8+y}{9+y}}} &=& \frac{1}{1 + \frac{1}{\frac{4y+35}{9+y}}}\\\\ &=& \frac{1}{1+\frac{9+y}{4y+35}} &=& \frac{1}{\frac{5y+44}{4y+35}}\\\\ &=& \frac{4y+35}{5y+44} && \end{array}$$

But then $$y = \frac{4y+35}{5y+44}$$ yields the quadratic $$5y^2+44y = 4y+35$$ which simplifies to $$y^2+8y-7 = 0$$ Finally, using the quadratic formula (and noting that $y \gt 0$) we have: $$y = -4 + \sqrt{23}$$ Thus, $$[4;\overline{1,3,1,8}] = 4 + y = \sqrt{23}$$